作为一个安可，我运行了一个排序网络生成器的长度10并将代码（使用选项“best”生成）移植到 Prolog/clpfd。
来了list_sorted__SN10/2（SN10代表“排序网络大小 10”）：
:- use_module(library(clpfd)).
list_sorted__SN10(Xs,Zs) :-
Xs = [A0,A1,A2,A3,A4,A5,A6,A7,A8,A9],
Zs = [E0,G1,H2,I3,J4,J5,I6,H7,G8,E9],
B4 #= min(A4,A9), B9 #= max(A4,A9),
B3 #= min(A3,A8), B8 #= max(A3,A8),
B2 #= min(A2,A7), B7 #= max(A2,A7),
B1 #= min(A1,A6), B6 #= max(A1,A6),
B0 #= min(A0,A5), B5 #= max(A0,A5),
C1 #= min(B1,B4), C4 #= max(B1,B4),
C6 #= min(B6,B9), C9 #= max(B6,B9),
C0 #= min(B0,B3), C3 #= max(B0,B3),
C5 #= min(B5,B8), C8 #= max(B5,B8),
D0 #= min(C0,B2), D2 #= max(C0,B2),
D3 #= min(C3,C6), D6 #= max(C3,C6),
D7 #= min(B7,C9), D9 #= max(B7,C9),
E0 #= min(D0,C1), E1 #= max(D0,C1),
E2 #= min(D2,C4), E4 #= max(D2,C4),
E5 #= min(C5,D7), E7 #= max(C5,D7),
E8 #= min(C8,D9), E9 #= max(C8,D9),
F1 #= min(E1,E2), F2 #= max(E1,E2),
F4 #= min(E4,D6), F6 #= max(E4,D6),
F7 #= min(E7,E8), F8 #= max(E7,E8),
F3 #= min(D3,E5), F5 #= max(D3,E5),
G2 #= min(F2,F5), G5 #= max(F2,F5),
G6 #= min(F6,F8), G8 #= max(F6,F8),
G1 #= min(F1,F3), G3 #= max(F1,F3),
G4 #= min(F4,F7), G7 #= max(F4,F7),
H2 #= min(G2,G3), H3 #= max(G2,G3),
H6 #= min(G6,G7), H7 #= max(G6,G7),
I3 #= min(H3,G4), I4 #= max(H3,G4),
I5 #= min(G5,H6), I6 #= max(G5,H6),
J4 #= min(I4,I5), J5 #= max(I4,I5).
让我们看看它是否有效：
?- numlist(1,10,Xs),permutation(Xs,Ys),\+ list_sorted__SN10(Ys,Xs).
false. % all permutations are sorted correctly
往另一个方向走呢？
?- list_sorted__SN10(Xs,[1,2,3,4,5,6,7,8,9,10]),
labeling([],Xs),
write('Xs'=Xs),nl,
false.
Xs=[1,2,3,4,5,6,7,8,9,10]
Xs=[1,2,3,4,5,6,7,8,10,9]
Xs=[1,2,3,4,5,6,7,9,8,10]
Xs=[1,2,3,4,5,6,7,9,10,8]
Xs=[1,2,3,4,5,6,7,10,8,9]
Xs=[1,2,3,4,5,6,7,10,9,8]
Xs=[1,2,3,4,5,6,8,7,9,10]
...
有速度吗？
?- time(list_sorted__SN10([10,9,8,7,6,5,4,3,2,1],Xs)).
% 198 inferences, 0.000 CPU in 0.000 seconds (97% CPU, 4841431 Lips)
Xs = [1, 2, 3, 4, 5, 6, 7, 8, 9|...].
有速度！
处理一般情况
排序列表很好，但如果我Xs有length(Xs,10)更长或更短的列表怎么办？
再次，对网络进行分类以进行救援！
这是 Bitonic 排序网络中显示的代码的 Prolog/clpfd 端口，用于 n 不是 2 的幂；Prolog 代码使用属性变量对要排序的项目进行随机读/写访问。我们使用一个属性value将项目存储在当时的特定位置。
:- use_module(library(clpfd)).
init_att_var(X,Z) :-
put_attr(Z,value,X).
get_att_value(Var,Value) :-
get_attr(Var,value,Value).
direction_flipped(ascending,descending).
direction_flipped(descending,ascending).
fdBitonicSort(Xs0,Zs) :-
same_length(Xs0,Zs),
maplist(init_att_var,Xs0,Xs1),
Xs2 =.. [data|Xs1],
functor(Xs2,_,N),
fdBitonicSort_(Xs2,0,N,ascending),
maplist(get_att_value,Xs1,Zs).
双调排序所需的递归分解由以下代码完成：
fdBitonicSort_(Data,Lo,N,Dir) :-
( N > 1
-> M is N // 2,
direction_flipped(Dir,Dir1),
fdBitonicSort_(Data,Lo,M,Dir1),
Lo1 is Lo + M,
N1 is N - M,
fdBitonicSort_(Data,Lo1,N1,Dir),
fdBitonicMerge_(Data,Lo,N,Dir)
; true
).
greatestPowerOfTwoLessThan(N,K) :-
T is 1 << msb(N),
( N /\ (N-1) =:= 0
-> K is T >> 1
; K = T
).
fdBitonicMerge_(Data,Lo,N,Dir) :-
( N > 1
-> greatestPowerOfTwoLessThan(N,M),
Ub is Lo + N - M,
fdBitonicCompareMany_(Data,Lo,Ub,M,Dir),
fdBitonicMerge_(Data,Lo,M,Dir),
Lo1 is Lo + M,
N1 is N - M,
fdBitonicMerge_(Data,Lo1,N1,Dir)
; true
).
比较的内部循环如下所示：
fdBitonicCompareMany_(Data,I,Ub,M,Dir) :-
( I < Ub
-> I_plus_M is I+M,
fdBitonicCompareTwo_(Data,I,I_plus_M,Dir),
I1 is I + 1,
fdBitonicCompareMany_(Data,I1,Ub,M,Dir)
; true
).
快完成了！缺少一件事...fdBitonicCompareTwo_/4读取第 i 个和第 j 个项目，如果方向是 ，则将最小值和最大值放在第 i 个和第 j 个位置ascending。如果方向是descending最小值和最大值，则放在第 j 和第 i 位：
fdBitonicCompareTwo_(Data,I,J,Dir) :-
I1 is I+1,
J1 is J+1,
arg(I1,Data,V1),
arg(J1,Data,V2),
get_attr(V1,value,W1),
get_attr(V2,value,W2),
Z1 #= min(W1,W2),
Z2 #= max(W1,W2),
( Dir == ascending
-> E1 = Z1, E2 = Z2
; E1 = Z2, E2 = Z1
),
put_attr(V1,value,E1),
put_attr(V2,value,E2).
测试
首先，对每个列表长度在1and之间200取随机数10 次，1并对10000它们进行排序。msort/2如果结果与交付的不同，请大声尖叫。
?- ( setrand(rand(29989,9973,997)),
between(1,200,N),
length(Xs,N),
format('(~d)',[N]),
( N mod 10 =:= 0 -> nl ; true ),
between(1,10,_),
maplist(random_between(1,10000),Xs),
( fdBitonicSort(Xs,Zs), \+ msort(Xs,Zs)
-> write(error(Xs,Zs)), nl
; true
),
false
; true
).
(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)
(11)(12)(13)(14)(15)(16)(17)(18)(19)(20)
(21)(22)(23)(24)(25)(26)(27)(28)(29)(30)
(31)(32)(33)(34)(35)(36)(37)(38)(39)(40)
(41)(42)(43)(44)(45)(46)(47)(48)(49)(50)
(51)(52)(53)(54)(55)(56)(57)(58)(59)(60)
(61)(62)(63)(64)(65)(66)(67)(68)(69)(70)
(71)(72)(73)(74)(75)(76)(77)(78)(79)(80)
(81)(82)(83)(84)(85)(86)(87)(88)(89)(90)
(91)(92)(93)(94)(95)(96)(97)(98)(99)(100)
(101)(102)(103)(104)(105)(106)(107)(108)(109)(110)
(111)(112)(113)(114)(115)(116)(117)(118)(119)(120)
(121)(122)(123)(124)(125)(126)(127)(128)(129)(130)
(131)(132)(133)(134)(135)(136)(137)(138)(139)(140)
(141)(142)(143)(144)(145)(146)(147)(148)(149)(150)
(151)(152)(153)(154)(155)(156)(157)(158)(159)(160)
(161)(162)(163)(164)(165)(166)(167)(168)(169)(170)
(171)(172)(173)(174)(175)(176)(177)(178)(179)(180)
(181)(182)(183)(184)(185)(186)(187)(188)(189)(190)
(191)(192)(193)(194)(195)(196)(197)(198)(199)(200)
true.
接下来，从1to N(with N =< Ub) 取出列表，考虑所有排列并查看其中任何一个显示双调排序中的错误（结果与msort/2给我们的结果不同）。
测试以两种不同的方式完成：after和before。after建立约束网络，然后将 FD 变量绑定到具体值。
before反过来，有效地将 clpfd 用作整数算术——所有约束都立即得到解决。
test_fdBitonicSort(Method,Ub) :-
length(RefList,Ub),
append(Xs,_,RefList),
length(Xs,N),
numlist(1,N,Xs),
same_length(Xs,Ys),
same_length(Xs,Zs),
time((format('[~q] testing length ~d (all permutations of ~q) ... ',
[Method,N,Xs]),
( Method == before
-> ( permutation(Xs,Ys),
\+ fdBitonicSort(Ys,Xs)
-> write(errorB(Ys))
; true
)
; permutation(Xs,Ys),
\+ (fdBitonicSort(Zs,Xs), Zs = Ys)
-> write(errorA(Ys))
; true
),
write('DONE\n'))),
false.
test_fdBitonicSort(_,_).
让我们运行test_fdBitonicSort/2：
?- test_fdBitonicSort(after,7).
[after] testing length 1 (all permutations of [1]) ... DONE
% 93 inferences, 0.000 CPU in 0.000 seconds (89% CPU, 1620943 Lips)
[after] testing length 2 (all permutations of [1,2]) ... DONE
% 4,775 inferences, 0.001 CPU in 0.001 seconds (100% CPU, 9136675 Lips)
[after] testing length 3 (all permutations of [1,2,3]) ... DONE
% 53,739 inferences, 0.006 CPU in 0.006 seconds (100% CPU, 9514148 Lips)
[after] testing length 4 (all permutations of [1,2,3,4]) ... DONE
% 462,798 inferences, 0.048 CPU in 0.048 seconds (100% CPU, 9652164 Lips)
[after] testing length 5 (all permutations of [1,2,3,4,5]) ... DONE
% 3,618,226 inferences, 0.374 CPU in 0.374 seconds (100% CPU, 9666074 Lips)
[after] testing length 6 (all permutations of [1,2,3,4,5,6]) ... DONE
% 32,890,387 inferences, 3.212 CPU in 3.211 seconds (100% CPU, 10241324 Lips)
[after] testing length 7 (all permutations of [1,2,3,4,5,6,7]) ... DONE
% 330,442,005 inferences, 32.499 CPU in 32.493 seconds (100% CPU, 10167747 Lips)
true.
让我们再次使用谓词，这次是地面输入：
?- test_fdBitonicSort(before,9).
[before] testing length 1 (all permutations of [1]) ... DONE
% 27 inferences, 0.000 CPU in 0.000 seconds (97% CPU, 334208 Lips)
[before] testing length 2 (all permutations of [1,2]) ... DONE
% 151 inferences, 0.000 CPU in 0.000 seconds (100% CPU, 1824884 Lips)
[before] testing length 3 (all permutations of [1,2,3]) ... DONE
% 930 inferences, 0.000 CPU in 0.000 seconds (100% CPU, 4308089 Lips)
[before] testing length 4 (all permutations of [1,2,3,4]) ... DONE
% 6,033 inferences, 0.001 CPU in 0.001 seconds (100% CPU, 5124516 Lips)
[before] testing length 5 (all permutations of [1,2,3,4,5]) ... DONE
% 43,584 inferences, 0.006 CPU in 0.006 seconds (100% CPU, 7722860 Lips)
[before] testing length 6 (all permutations of [1,2,3,4,5,6]) ... DONE
% 353,637 inferences, 0.033 CPU in 0.033 seconds (100% CPU, 10753040 Lips)
[before] testing length 7 (all permutations of [1,2,3,4,5,6,7]) ... DONE
% 3,201,186 inferences, 0.249 CPU in 0.249 seconds (100% CPU, 12844003 Lips)
[before] testing length 8 (all permutations of [1,2,3,4,5,6,7,8]) ... DONE
% 32,060,649 inferences, 2.595 CPU in 2.594 seconds (100% CPU, 12355290 Lips)
[before] testing length 9 (all permutations of [1,2,3,4,5,6,7,8,9]) ... DONE
% 340,437,636 inferences, 27.549 CPU in 27.541 seconds (100% CPU, 12357591 Lips)
true.
有用！还有更多事情要做吗？是的，绝对！
list_sorted__SN10/2首先，应该为其他小尺寸生成专门的代码。其次，可以评估不同的等效网络排序方法。